Skip to main content
OverflowAI is here! AI power for your Stack Overflow for Teams knowledge community. Learn more
15 votes

What is a Haar random quantum state?

Typically this is a slight abuse of notation. One can have a unitary operator $U$ chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state $|\psi_0\rangle$...
Quantum Mechanic's user avatar
10 votes
Accepted

Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

This answer won't actually give you a bound, but will provide some information that may help you in your search. You may be able to find an answer in the random matrix theory literature if you ...
John Watrous's user avatar
  • 6,087
9 votes

Expected value of a Haar random quantum state multiplied by a unitary

I'm writing an alternate proof because it uses some interesting tools, computes the value of these expressions, and gives some insights into how we can interpret the quantities in consideration. The ...
keisuke.akira's user avatar
8 votes
Accepted

Multiplication by a Haar random unitary two times

There is an explicit formula for the integral with respect to the Haar measure of any polynomial in the entries of a unitary and its conjugate, due to Collins and Śniady: Benoît Collins and Piotr ...
John Watrous's user avatar
  • 6,087
7 votes
Accepted

Confusion about the output distribution of Haar random quantum states

The two facts are connected in that they both arise as a result of rotational invariance of the Haar measure. We will derive them in the case of large $n$ since this is when the Porter-Thomas ...
Adam Zalcman's user avatar
  • 22.9k
7 votes
Accepted

Random quantum states and Schur-Weyl duality

Note that the quoted relation $$ \bar M_i = \sum_\lambda a_\lambda P_\lambda, $$ only holds if the $M_i$ also commute with the representation of the symmetric group! Otherwise this can obviously not ...
Markus Heinrich's user avatar
7 votes

What is the expectation value ${\Bbb E}[\langle\psi,O\psi\rangle]$ over the Haar distribution?

Since a Haar-random $\lvert\psi\rangle=U\lvert0\rangle$ for a Haar-random $U$, your expectation value equals $$ \langle 0 \rvert \Big[\int \mathrm d U\, UOU^\dagger\Big]\lvert0\rangle\ . $$ The ...
Norbert Schuch's user avatar
6 votes

How to show that the integral over all Haar states vanishes: $\int|\psi\rangle\,{\rm d}\psi = 0 $?

Yes, we can show this using the unitary invariance of the Haar measure on states. In more detail, we have $$ U \int |\psi\rangle\, \mathrm{d}\psi = \int U|\psi\rangle\, \mathrm{d}\psi = \int |\psi\...
Markus Heinrich's user avatar
6 votes

Is the Haar measure invariant under conjugation?

I will answer this question in a more general context. You might know that Haar's theorem tells you that on any locally compact group $G$, there is a unique left-invariant (Borel) measure $\mu$, up to ...
Markus Heinrich's user avatar
6 votes
Accepted

How close or far apart are the distributions generated by two Haar random states?

Since the Haar-measure is unitarily invariant, the $\mathbf{D}_\psi$ that we obtain will be independent of $\psi$. In fact, the $\mathbf{D}_\psi$ obtained from measuring $\psi$ with respect to any ...
keisuke.akira's user avatar
5 votes
Accepted

Compute the large $n$ distribution of $|\langle z_i|\psi\rangle|^2$ over Haar random quantum states

In the following, I'll show the evaluation of the probability densities of the transition probabilities: $|\langle \psi | z\rangle^2$ and their pairwise independence. I didn't work out the full mutual ...
David Bar Moshe's user avatar
5 votes
Accepted

Expected value of a Haar random quantum state multiplied by a unitary

With the chosen structure of $ U $, i think it's even possible to prove the stronger statement: $$ \langle z| \rho|z \rangle = \langle z| \sigma_\rho|z \rangle, \hspace{0.2em} \text{where} \hspace{0....
tsgeorgios's user avatar
  • 1,416
5 votes
Accepted

Computing expectation value of $|\langle z|C|0^n\rangle|^2$ over Haar random circuit

The issue that easily leads to confusion is the dual role played by output bitstring probability. It enters the computation of the average in two ways. On one hand, it determines how often one sees ...
Adam Zalcman's user avatar
  • 22.9k
5 votes

Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

The result you're looking for is effectively Proposition 19 of the paper: Almost all quantum channels are equidistant; which I'm rewriting here for convenience: Let $U, V \in \mathcal{U}(d)$ be two ...
keisuke.akira's user avatar
5 votes
Accepted

Generating random, but non-uniform state

Rejection sampling is a good fit and works without any changes, simply by plugging the desired distribution $p(\psi)$ into the standard algorithm. Let$^1$ $M:=\max_{\psi\in\mathbb{CP}^1} p(\psi)$. To ...
Adam Zalcman's user avatar
  • 22.9k
5 votes
Accepted

What is the expectation value of $|\langle \psi|U|\psi \rangle|$ over Haar random states $|\psi\rangle$?

This partial answer calculates the integral for $d=2$. In this case, every traceless unitary $U$ is equivalent to the Pauli $Z$ up to similarity and global phase, so, by rotational invariance of the ...
Adam Zalcman's user avatar
  • 22.9k
5 votes
Accepted

Is there a concentration inequality for the quantum gate fidelity $F(C,U)$ for a channel $C$ such that $\int dU F(C,U)=X$?

The function described in the question is 1-Lipschitz. To argue this, we'll get an inequality in place before we start writing integrals. If $\vert \gamma\rangle$ and $\vert\delta\rangle$ are unit ...
John Watrous's user avatar
  • 6,087
4 votes

Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state. That means that there is some state, call it $|\psi\rangle$, ...
glS's user avatar
  • 25.2k
4 votes

Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Yes. A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as ...
Norbert Schuch's user avatar
4 votes

On the distribution of the fidelity of a random product state with an arbitrary many-qubit state

All you need are simple tools from measure concentration. The setup is as follows (repeated from the question above for completeness): $| \psi \rangle$ is an $n$-qubit state and $| \alpha \rangle := | ...
keisuke.akira's user avatar
4 votes
Accepted

At what depth and for what architecture are random quantum circuits $1$-designs?

To study unitary $t$-designs, we define the moment operator with respect to a probability measure $\nu$ as $$ M_t(\nu) := \int_{U(d)} U^{\otimes t} (\cdot) (U^{\otimes t})^\dagger d\nu(U) \simeq \int_{...
Markus Heinrich's user avatar
4 votes
Accepted

Anticoncentration for two independent random quantum circuits in parallel

I believe so (caveat: this is not something I've every thought about before). I'm going to rewrite the $p_x$ from your question as $p_{xy}$. So, we have $$ p_{xy}=|\langle x|U_1|0^n\rangle|^2\ |\...
DaftWullie's user avatar
  • 58.7k
4 votes
Accepted

Moments of Pauli coefficients of Haar-random states

Let us compute the value for $\alpha=4$, averaged over Haar-random states. We have the following identity: $$ \sum_{P\in\mathcal{P}_n} \mathrm{tr}(\rho P)^4= \sum_{P\in\mathcal{P}_n} \mathrm{tr}(\rho^...
Markus Heinrich's user avatar
3 votes

Spoofing XQUATH with the Feynman method

The paper does not specify the exact algorithm or class of distributions $\mathcal{D}$ for which such algorithm fails to refute XQUATH, and some classes of distributions $\mathcal{D}$ do not satisfy ...
fiktor's user avatar
  • 326
3 votes

What is the probability $\Pr(\|U-I\|_{\rm op}<\varepsilon)$ of a Haar-random unitary being close to the identity?

$U-I$ is a normal matrix so $||U-I||_{op}$ is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\...
Quantum Mechanic's user avatar
3 votes

On the distribution of the fidelity of a random product state with an arbitrary many-qubit state

The required fidelity $F$ is a function of the Cartesian product of the single $n$-qubit state space: $CP^{2^n-1}$ and $n$ copies of a single qubit state space: $CP^{1} \cong S^2$. The statistics ...
David Bar Moshe's user avatar
3 votes
Accepted

Quantum hardness of XQUATH conjecture

Maybe think of it this way - a quantum computer, executing a small enough random circuit $C$ acting on a state initially prepared as $\vert 0^n\rangle$ and sampling therefrom, will get an $n$-bit ...
Mark Spinelli's user avatar
3 votes
Accepted

Quantum supremacy: shallow depth Haar random circuits and unitary designs

First of all, that does not imply anything for shorter (constant/logarithmic) depths. Moreover, the 2-design property does not imply that the outcome distribution is the same as for Haar-random ...
Markus Heinrich's user avatar
3 votes

Prove that uniformly random states have moments ${\bf E}_\psi|\langle x|\psi\rangle|^{2t}\sim1/\binom d t$

The factor in your claim is wrong. It should be $\binom{d+t-1}{t}^{-1}$. The correct claim follows from the identity $$ \int |\psi\rangle\langle\psi|^{\otimes t} d\psi = \binom{d+t-1}{t}^{-1} P_{\...
Markus Heinrich's user avatar
3 votes
Accepted

Density matrices of multiples copies of a single Haar-Random state

You can think of this as a generalization of the maybe better-known result that $$\int d\psi \,\mathbb{P}_\psi = \frac{I}{d},$$ where I used the notation $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\...
glS's user avatar
  • 25.2k

Only top scored, non community-wiki answers of a minimum length are eligible